- anomer of an aldohexose E has a specific rotation of 120- while - anomer of E has a specific rotation of 80- If an equilibrium mixture of - - anomers of E has the specific rotation of 96- then what will be percentage of the -anomer in the equilibrium mixture assuming the concentration of the open structure is almost negligible-

Let, a be the mole fraction of nbsp;α anomer and b nbsp;is the mole fraction of nbsp; β anomer. nbsp; Then, a+b=1 120^∘a+80^∘(1 a)=96 ⇒ a=0.4 nbsp; So ...

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Standard XII

Chemistry

Liquids in Liquids

α anomer of a...

Question

$α$ anomer of an aldohexose E has a specific rotation of $120^{\circ}$ while $β$ anomer of E has a specific rotation of $80^{\circ}$ . If an equilibrium mixture of $α, β$ anomers of E has the specific rotation of $96^{\circ}$ , then what will be percentage of the $α$ -anomer in the equilibrium mixture assuming the concentration of the open structure is almost negligible?

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Solution

Let, a be the mole fraction of $α$ anomer and b is the mole fraction of $β$ anomer.
Then, $a + b = 1$
$120^{\circ} a + 80^{\circ} (1 - a) = 96$
$\Rightarrow a = 0.4$
So the percentage of $α$ -anomer in the equilibrium mixture assuming the concentration of the open structure is almost negligible will be $0.4 \times 100 = 40 %$

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Let, a be the mole fraction of α anomer and b is the mole fraction of β anomer. Then, a+b=1 120∘a+80∘(1−a)=96 ⇒a=0.4 So the percentage of α-anomer in the equilibrium mixture assuming the concentration of the open structure is almost negligible will be 0.4×100=40%